Elec467 Power Machines

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Elec467 Power Machines Transformers

• Electric Machines by Hubert, Chapter 2
• Topics Exciting current, Ideal transformer,
  Equivalent circuits, Per Unit calculation,
• Open circuit/short circuit testing models

2
Typical physical designs for a transformer
3
Basic transformer action
ex is the counter emf
(Formula 2-3) Voltage ratio turns ratio
4
Calculating transformer flux

• From Example 2.1
• Given that a transformer 50-kVA, 200 turns, 240V,
  60 Hz primary, what is the maximum flux, Fmax?
• Use formula 2.1 Ep 4.44 Np f Fmax
• Insert knowns 240 4.4420060 F
• Solve for the unknown
• Fmax 240/(4.4420060)?.0045 Wb

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Section 2-5No load conditions
In this diagram is a simplified view of the
primary side of a transformer. The input current
IP I0 (the exciting current) during no load
conditions which can be measured directly with an
ampere meter. We see the eddy loss component
(Ife) caused by the resistance Rfe (a fictitious
resistance to account for core loss) and the
magnetizing current (IM) present during no load
conditions whose reactance component is jXM is in
parallel. The value jXM is a fictitious
reactance to account for the magnetizing current.
Their vector summation gives the current input
when there is no load I0 Ife IM. Additional
equations from formula 2-4 Ife VT/Rfe IM
VT/jXM
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Example 2.3Calculating flux loss for no-load
conditions

• The exciting circuit, Io , is a no-load current.
  Lets calculate the exciting circuit in the
  primary of a transformer using data from Example
  2.3
• Given a 25-kVA, 2400240 V, 60 Hz, draws 138 W at
  no-load condition with a .210 lagging power
  factor (current into a coil lags the voltage).
• 1st get the phase angle ?cos-1(.210) 77.88
• 2nd insert the power factor angle of 77.88 into
  the formula for the phase difference between the
  voltage and current ? ?v ?i
• Using 0 phase for ?v gives 77.88 ?i
  therefore ?i 77.88
• This value is the angle the current vector I0
  makes with the vector Ife
• Well need this information to calculate the IM
  current but first lets do the core-loss
  component that draws real power Pcore loss Vt
  Ife
• Insert known values 138 2400 Ife
• Solve for the unknown Ife .0575 amps giving
  us the amplitude for the vector Ife
• Utilize this value and a trig formula from the
  power triangle cos ?i Ife / Io
• Insert known values .210 .0575/ Io yields up
  Io .2738 amps
• Using Io Ife IM we can solve for IM Io -
  Ife ? .2738 - .0575 .268 amps
• Summary Io .2738 amps, Ife .0575 amps, IM
  .268 amps

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Formulas used for no load calculations

• FM the mutual flux produced by the magnetizing
  component of the exciting current
• NpIM is aka magnomotive force, F
• VT IpRp Ep
• Ip (VT Ep)/Rp
• Above formulas are 2-6, 2-7 2-8 from text, page
  46.

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Formulas used for calculation of quadrature
components

• Vt Ife Rfe multiply both sides by Ife and we
  get real power loss in watts equal to I2R
• Pcore loss Vt Ife
• Vt IM j XM reaction power loss in watts
• Io Ife IM (this is a vector formula)
• The vector diagram
• from Fig. 2-4 ?

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A closer look at the phasors

• Figure 2-4 phasor drawing is seen at the right.
  ?i is the power factor angle. When rearranged as
  seen on the right
• cosine adj/hyp?Ife/Io
• giving us one more handy formula
• Io Ife/cos ?i or
• Io Ife/power factor

Io
IM
?
Ife
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Mutual flux
When a load is placed on the secondary the
primary side changes from just the exciting
current (Io Ife IM) to the exciting current
plus a load current, IP I0 IP,load (formula
2-11).
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Section 2-7Flux loss
Flux leakage results in a different induced
voltage in the secondary voltage than is no
leakage occurred. It diminishes the efficiency
of the transformer. Flux leakage is accounted for
in the formulas Fp FM Flp formula 2-12
Fs FM Fls formula 2-13
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Section 2-8Ideal transformer
The ideal transformer is represented above. The
induced counter-emf voltages and the input
impedance are designated with primed symbols.
There is a long list of what is not included in
the ideal transformer. When the turns ratio is
not available use the nameplate voltage ratio.
The following slide lists a number of formulas
used with an ideal transformer. The effects of
flux leakage and winding resistance are
insignificant at no load.
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Ideal transformer formulas
Assuming the primary is the High Side
Input Impedance of an Ideal Transformer
since
The apparent power input must equal the apparent
power output
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Equivalent circuit parameters for a real
transformer
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Close up look at the ideal transformer from Fig.
2-8
In the ideal Ep 4.44Np f Fp and Es 4.44Ns f
Fs If the windings have the same polarity, the
current will appear to flow into the transformer
and flow out the same end as if the primary and
secondary circuits were connected to together.
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Equivalent circuit details
An equivalent circuit of a real transformer using
an ideal transformer and external components to
represent what is going on internal to the
transformer. Losses are represented by IPRP,
Elp, IfeRfe and by the magnetizing current IMjXM.

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Leakage Reactance
When we include the flux leakage (formulas 2-12
13) in the ideals equations we get Ep 4.44Np
f FM 4.44Np f Flp formula 2-20 Es 4.44Ns f
FM - 4.44Ns f F ls formula 2-21
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Formulas we need to know to calculate voltages
around the loops

• Expressing formulas 2-20 21 in simpler form
• EP EP Elp ES ES - Els
• VT Ep IpRp VT E lp Ep IpRp
• Ip Ife IM Ip load formula 2-26
• Es IsRs Vs load
• Es Els IsRs Vs load formula 2-28
• The sinusoidal voltage generated
• E Emaxcos(2pft) with Emax 2pfNFmax formula
  2-30
• L N2/R formula 2-32

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Equivalent impedance usingreferred parameters
20
Referring 2nd parameters to the primary side
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Formulas used for Zin

• Ideal transformer Zin a2Es/Is (see Fig.
  2-9)
• Where Is Es/(Rs jXls Zload)
• Substituting Is into the first equation gives
• Zin a2(Rs jXls Zload) ?
• Zin a2Rs a2jXls a2Zload
• The impedance seen by the source is
• Zin Zin (RP jXlP)

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How to calculate the core losses seen by the
primary

• Neglect exciting current at near rated load
  conditionsthe equivalent impedance is
• ZeqP Rp a2Rs j(Xlp a2Xls)
• and, Zin, HS ZeqP a2Zload (note this is
  not Zin)

23
Equivalent impedance HS is defined for a
step-down transformer
Here they have taken the impedance in Rp a2Rs
j(Xlp a2Xls) from the previous slide and
changed it into Zeq,HS Req,HS jXeq,HS by
grouping the resistances and impedance together
Req,HS RP a2RS jXeq,HS j(XlP a2XlS)
(note Req,HS jXeq,HS are the parameters
obtained in the short-circuit test)
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Taking the core losses to the secondary
Notice that the loads when reflected thru the
transformer use a2 but current and voltage
relations only use a. Current as usual is
inverted.
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How to calculate the wiring and flux losses in
the secondary

• ZeqS Rs Rp/a2 j(Xls Xlp/a2)
• Here the primary values are adjusted by 1/a2
  which is the result of going the other direction
  across the turns ratio.

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Transformer as a lens

• Think of the transformer as a telescope.
• Telescopes make thing look larger or smaller
  depending on which way you look thru it.
• When the high voltage is stepped down, going from
  primary to secondary make values smaller
• Thus the primary losses when taken to the
  secondary are divided by a2.
• But bringing the secondary values to the primary
  side makes them bigger thus Zload and the
  secondary losses are multiplied by a2.

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Equivalent circuit for a step-down
transformerlow side
The secondary impedance is changed from Rs
Rp/a2 j(Xls Xlp/a2) seen in the last slide
into Zeq,LS Req,LS jXeq,LS by grouping the
resistances impedances together Req,LS RLS
RHS/a2 jXeqLS j(XLS XHS/a2)
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Equivalent circuit for a step-up transformer can
go either way also
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Changing from a Step-Down to a Step-Up

• If we think of the inputs and output of a
  transformer as primary (the driving circuit) and
  secondary (the load) an adjustment to a is
  necessary when changing from a step-down to a
  step-up or visa-versa.
• The value of a is inverted because the turns
  ratio is inverted.
• We can avoid this problem by using formulas that
  refers to the opposite sides of the transformer
  in terms of voltage high side or low side.

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Equivalent circuit for a step-up transformerlow
side
For the step-up mode the source is on the low
side (smaller number of windings). The
equivalent impedance is Zeq,LS RLSRHS/a2
j(XlLS XlHS/a2) or Zeq,LS Req,LS jXeq,LS
where Req,LS gather the real terms and jXeq,LS
gathers the imaginary terms. Zin,LS (1/a2
)Zload,HS Zeq,LS
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Equivalent circuit for a step-up transformerhigh
side
The equivalent impedance on the high side uses
Zeq,HS RHS a2RLS j(XlHS a2XlLS). The
impedance show above is Zeq,HS Req,HS
jXeq,HS where Req,HS gathers the real terms and
jXeq,HS gathers the imaginary terms.
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Voltage Regulation

• Regulation (Enl Vrated)/ Vrated
• Enl is the voltage present at the output
  terminals when no load is present. Vrated is
  the name plate voltage.
• Enl ILSZeq,LS VLS formula 2-45 (p. 62)
• ILS Papparent/VLS consider VLS as Vrated
• Zeq,LS Zeq,HS/a2 formula 2-43a (p. 59)

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Per Unit Parameters

• Use by professionals in the power industry, per
  unit parameters simplify voltage calculation for
  transmission lines.
• ZPU Irated Zeq)/Vrated
• RPU Irated Req)/Vrated
• XPU Irated Xeq)/Vrated
• Zbase Vrated/Irated also V2rated/Srated
• Phase angle a tan-1(XPU/RPU)

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Transformer losses and efficiency

• ? Pout/Pin
• ? Pout/(Pout Pcore I2Req)
• Pout is the apparent power (xxKVA)
• Pcore is the wattage reading showing the real
  power lost under no load conditions. This
  happens to be the (HS) value of the power reading
  taken during the open circuit test!
• I2Req is the wattage reading taken during the
  short circuit test!
• Therefore ? Pout/(Pout POC PSC)

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Open Circuit Test
Open circuit test is measured on the low side
with the high side open so that only the exciting
current parameters are detected.
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Formulas used in the Open Circuit Test

• Parameters measure during an open circuit test
  are POC, VOC, IOC
• POC VOCIfe ? Ife POC/VOC
• IOC v(Ife2 IM2) ? IM v(IOC2 Ife2)
• Rfe,LS VOC/Ife
• XM,LS VOC/IM
• Note By conducting an open circuit test from
  both sides, Rfe and XM can be determined for both
  primary and secondary sides. These values are
  used in no-load calculations in step-down and
  step-up transformers.

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Short circuit test
Short-circuit test detects the high side
equivalent parameters as there is no-load seen
and the exciting current is ignored.
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Formulas for the short circuit test

• Parameters measure during a short circuit test
  are PSC, VSC, ISC
• ISC VSC/Zeq,HS ? Zeq,HS VSC/ISC
• PSC ISC2 Req,HS ? Req,HS PSC/ISC2
• Zeq,HS v(Req,HS2 Xeq,HS2) ?
• Xeq,HS v(Zeq,HS2 - Req,HS2)

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Homework Assignment

• 2-3
• 2-5
• 2-9
• 2-11
• 2-13
• 2-17 a b
• 2-27
• 2-34 a
• 2-39 a